Actually, Amber can win even with $2n$ piles. At every turn, she looks at all $k$ such that there are at least two piles of size $k$, and chooses the largest such $k$, say $k_0$. Then she queries two piles of the chosen size $k_0$. Clearly Barbara has only one move: make the piles of size $k_0+1$ and $k_0-1$. It is easy to see that this process terminates; in particular the value of $k_0$ must decrease after a chain of moves. At the end the piles all have distinct sizes. If there is no empty pile then the total number of stones is at least $$1+2+ \cdots 2n = n(2n+1) > 2n \cdot n$$Contradiction! However I still haven't proved that $2n$ is the lowest...

I think I managed to prove the other part. Suppose there are $2n-1$ piles. Barbara uses the following strategy: If the piles are unequal , then move a stone from the larger pile to the smaller pile (if the piles are equal there is only one move). Call a move reducing if the two piles are unequal, and firing otherwise. We denote every move by $(i,j)$ where $i,j$ are the sizes of the two piles queried. Assume FTSOC that Amber somehow manages to create an empty pile. Choose the smallest possible sequence of moves that can achieve this goal. Claim: There are no reducing moves. Proof: Assume the contrary, and let the final reducing move be $(i,j)$ where $i>j$. If neither $(i-1,i-1)$ nor $(j+1,j+1)$ are played afterwards, then we can delete the move $(i,j)$, contradicting minimality. Else WLOG $(j+1,j+1)$ happens first. Then we can delete $(i,j)$ and $(j+1,j+1)$, and replace them by the single move $(i,j+1)$, contradiction! $\blacksquare$ Now we imagine the game in a different way. Consider the integer number line. At every point $k$, we place as many chips there as there as piles of size $k$ (we hypothetically allow piles to have negative size too). Since there are no reducing moves, the game becomes an incomplete version of Spencer's chip firing game, but with initial condition translated by $n$ to the right. In particular, we use the following result from the above paper: Every such game ends at the same final configuration, regardless of moves, such that there is one chip each at $1,2, \dots 2n-1$ (Theorem 3.9). But by our initial assumption, there is a chip at $0$ at some point in the game, so after that, if we continue the game arbitrarily, there would always be a chip at some non-positive number. But this contradicts the final configuration! $\blacksquare$

Nice solution to the second part. My original solution is like this: Construct a complete graph on $4039$ vertices, we will add an orientation to each edge that we decide later, such that if we assigned a directed edge $i\rightarrow j$, then if Amber picks two piles $i$ and $j$ for an odd number of time, Barbara move one stone from $i$ to $j$ following that orientation, otherwise she move one stone from $j$ to $i$ instead. Till this end it suffices to construct an orientation of the edges on the complete graph such that each vertex has outdegree of at most $2019$. We appeal to the following lemma: $\textbf{Lemma.}$ Any complete graph of $2m+1$ vertices can be partitioned into disjoint cycles of length $2m+1$ each. $\textit{Proof.}$ Label the vertices as $v, u_1, u_2, \cdots u_{2m}$, then for $1\le i\le m$, we can construct a cycle of the form $\cdots - u_i - v - u_{i+m} - \cdots$, by a ''zig-zag'' construction: $$u_i, u_{i+1}, u_{i-1}, u_{i+2}, u_{i-2}, \cdots, u_{i+k}, u_{i-k}, \cdots, u_{i+m}, v, u_i$$This partitions the complete graph into $m$ cycles above for $1\le i\le m$. $\square$ Finally, take all these cycles and make them a directed cycle each. This resulting graph will have each vertex has outdegree of exactly $2019$, and we are done. $\blacksquare$.

navi_09220114 in post#5 wrote: Suppose that Amber have $2k+2$ piles of $2(k+1)^2$ rocks each, then pick the first $2k$ piles, each with $k+1$ rocks. Why is it possible to assume so? Your generalized statement does not guaranteed that each pile has $m$ rocks.